imaginary numbers

• Mar 14th 2010, 07:02 PM
Anemori
imaginary numbers
Hello! I don't know how to foil and add/subtract an imaginary numbers.

$x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))$

thanks...
• Mar 14th 2010, 07:27 PM
skeeter
Quote:

Originally Posted by Anemori
Hello! I don't know how to foil and add/subtract an imaginary numbers.

$x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))$

thanks...

$x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]$

$x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

note that $i^2 = -1$ ... finish.
• Mar 14th 2010, 08:03 PM
Anemori
Quote:

Originally Posted by skeeter
$x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]$

$x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

note that $i^2 = -1$ ... finish.

$x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

$x^2(x^2+4)(x^2-6x+10)$

$x^2(x^4 - 6x^3+14x^2-24x+40)$

$x^6 - 6x^5+14x^4-24x^3+40x^2$

This is what i get... i hope this is right...
• Mar 14th 2010, 10:29 PM
Anemori
Quote:

Originally Posted by Anemori
$x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

$x^2(x^2+4)(x^2-6x+10)$

$x^2(x^4 - 6x^3+14x^2-24x+40)$

$x^6 - 6x^5+14x^4-24x^3+40x^2$

This is what i get... i hope this is right...

I dont know if this is right...

when i plug in back the factor from this polynomial, it doesn't go to zero.
• Mar 15th 2010, 11:45 AM
skeeter
Quote:

Originally Posted by Anemori
I dont know if this is right...

when i plug in back the factor from this polynomial, it doesn't go to zero.

the zeros are $x = \pm 2i$ , $x = 3 \pm i$ , and $x = 0$

... are these the values what you "plugged" in for x ?
• Mar 15th 2010, 01:56 PM
Anemori
Quote:

Originally Posted by skeeter
the zeros are $x = \pm 2i$ , $x = 3 \pm i$ , and $x = 0$

... are these the values what you "plugged" in for x ?

yes. but its not real zero. maybe im doing it wrong. can you show me one of them
• Mar 15th 2010, 02:14 PM
skeeter
Quote:

Originally Posted by Anemori
yes. but its not real zero. maybe im doing it wrong. can you show me one of them

$f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2
$

$f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$f(-2i) = -64 + 192i + 224 - 192i - 160 =0$
• Mar 15th 2010, 04:56 PM
Anemori
Quote:

Originally Posted by skeeter
$f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2
$

$f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$f(-2i) = -64 + 192i + 224 - 192i - 160 =0$

oh ok... maybe my calculator if messed up. im getting huge numbers heheheh. but it wouldnt make any difference if I use the exponents first before multiplying the outside paranthesis?
• Mar 16th 2010, 10:33 PM
Anemori
Quote:

Originally Posted by skeeter
$f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2
$

$f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$f(-2i) = -64 + 192i + 224 - 192i - 160 =0$

only -2i and 3-i will result of zero in the function. the rest are not zero.
• Mar 17th 2010, 07:29 AM
skeeter
Quote:

Originally Posted by Anemori
only -2i and 3-i will result of zero in the function. the rest are not zero.

is that so?

$f(2i) = (2i)^6 - 6(2i)^5 + 14(2i)^4 - 24(2i)^3 + 40(2i)^2
$

$f(2i) = 64i^6 - 192i^5 + 224i^4 - 192i^3 + 160i^2$

$f(2i) = -64 - 192i + 224 + 192i - 160 = 0$

... and I guarantee you that $f(3+i) = 0$ ... complex zeros always come in conjugate pairs.
• Mar 17th 2010, 08:15 AM
Chris L T521
There is enough information for the user to solve the problem. There is no need to carry on this discussion anymore.